ABC is a triangle right-angled at B. Points M and N trisect BC. If AM = 6 cm and AN = 9 cm, then what is AC equal to ?

Explanation

Let BM = x, so BN = 2x and BC = 3x. Let AB = c. By Pythagoras: c^2 + x^2 = 36 and c^2 + 4x^2 = 81. Subtracting the first from the second gives 3x^2 = 45, so x^2 = 15. Then c^2 = 36 - 15 = 21. AC^2 = c^2 + (3x)^2 = 21 + 9(15) = 156. Thus AC = \sqrt{156} = 2\sqrt{39} cm.

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